如何使用AngularJS中的$ http将文件发送到PHP脚本

I think my question is clear enough but here some details. I pretty new to Angular and PHP so I probably did a lot of mistakes or missing something important :).

我认为我的问题很清楚,但这里有一些细节。我对Angular和PHP很新,所以我可能犯了很多错误或遗漏了一些重要的东西:)。

I have first of all a form containing text and file inputs. Like this :

我首先拥有一个包含文本和文件输入的表单。像这样 :

<div><label for="textInput">Text input</label></div>
<div><input id="textInput" name="textInput" type="text" ng-model="form.textInput"></div><br/>
<div>Put some file please</div>
<div>
  <input id="file1" type="file" name="file1" ng-model="form.file1"><br/>
  <input id="file2" type="file" name="file2" ng-model="form.file2">
</div>

To post file with ng-model I used this directive :

要使用ng-model发布文件,我使用了这个指令:

(function () {
    fileInput.$inject = [];
    function fileInput() {
        var fileTypeRegex = /^file$/i;
        return {
            restrict: 'E',
            require: '?ngModel',
            link: link
        };
        function link(scope, element, attrs, ngModel) {
            if (ngModel && element[0].tagName === 'INPUT' && fileTypeRegex.test(attrs['type'])) {
                    element.on('change', function () {
                    var input = this;
                        if ('multiple' in attrs) {
                        var files = Array.prototype.map.call(input.files, function (file) { return file; });
                        ngModel.$setViewValue(files);
                    }
                   else {
                       ngModel.$setViewValue(input.files[0]);
                    }
                });
            }
        }
    }
    angular.module('ng-file-input', []).directive('input', fileInput);
}());

Finally, I send data with $http :

最后,我用$ http发送数据:

$http({
    url: window.API_URL + 'postulate.php',
    method:'POST',
    data:$scope.form
}).then(function successCallback(response){
    console.log(response.data);
    console.log($scope.form);
});

In PHP file I only get $_POST data and I print it :

在PHP文件中我只获得$ _POST数据并打印出来:

$rest_json = file_get_contents("php://input");
$_POST = json_decode($rest_json, true);
echo json_encode($_POST);

And here's the problem. I get with console.log() :

这就是问题所在。我得到console.log():

Object {textInput: "the content", file1: Array[0], file2: Array[0]}
Object {textInput: "the content", file1: File, file2: File}

Did a lot of Google and tests but I can't get it to work.

做了很多谷歌和测试,但我无法让它工作。

Ask me if you want more details.

问我是否想要更多细节。

1 个解决方案

#1

To get the files set trough ajax from PHP you need to use the $_FILES php global variable which actually is an array containing some field like name, type, tmp_name. So if you send multiple files, the combined array should look something like this:

要通过PHP中的ajax设置文件,您需要使用$ _FILES php全局变量,该变量实际上是一个包含名称,类型,tmp_name等字段的数组。因此,如果您发送多个文件,组合数组应如下所示:

 Array
(
    [image] => Array
        (
            [name] => MyFile1.jpg (comes from the browser, so treat as tainted)
            [type] => text/plain  (not sure where it gets this from - assume the browser, so treat as tainted)
            [tmp_name] => /tmp/php/php1h4j1o // this is were the temporally file is saved on the server
            [error] => UPLOAD_ERR_OK  (= 0)
            [size] => 123   (the size in bytes)
        )

    [image] => Array
        (
            [name] => MyFile2.jpg
            [type] => image/jpeg
            [tmp_name] => /tmp/php/php6hst32
            [error] => UPLOAD_ERR_OK
            [size] => 98174
        )
)

To obtain the files in php you can check if the $_FILES is set or it is not empty:

要获取php中的文件,您可以检查是否设置了$ _FILES或它是否为空:

if (isset($_FILES['image']['name']) && (file_exists($_FILES['image']['tmp_name']))) {
    // do the job here
}

Here the name of the image field depends on how you defined the form in your html. For example if you have an input field declared as:

这里图像字段的名称取决于您在html中定义表单的方式。例如,如果您将输入字段声明为:

<input type="file" name="image" id="image">

then the name here is how you identify the $_FILES first field. So actually you get the files based on the array key.

那么这里的名称是你如何识别$ _FILES第一个字段。实际上你实际上是根据数组键获取文件。

Hope that's clear.

希望很清楚。